Answer to Question #342780 in Statistics and Probability for elay

Question

An auto battery company claims that their batteries’ mean

life is 50 months. In order to check this claim, a DTI

researcher took a random sample of 18 of these batteries

and found that the mean life is 48.8 months with a

standard deviation of 7 months. Assume that battery life

follows a normal distribution, test with 90% confidence

whether the companies’ claim is different from the true

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=50$

$H_a:\mu\not=50$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.10,$ $df=n-1=17$ degrees of freedom, and the critical value for a two-tailed test is $t_c =1.739607.$The rejection region for this two-tailed test is $R = \{t:|t|>1.739607\}.$

The t-statistic is computed as follows:

$$t=\dfrac{48.8-50}{7/\sqrt{18}}=-0.7273$$

Since it is observed that $|t| = 0.7273<1.739607=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed $df=17$ degrees of freedom, $t=-0.7273$ is $p= 0.47694,$ and since $p= 0.47694>0.10=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 50, at the $\alpha = 0.10$ significance level.